Question

我似乎无法让我的代码100％运行。我试图让我的代码以这样的方式工作：

如果列值为空（$id = columnName），则插入该行。如果有一个＆＃34;值＆＃34;找到$id，然后更新已保存在数据库中的值。

功能：

     $socialCheck = SocialSettings::where($id)->first();

       $socialInsert = new SocialSettings;

       $value        = Request::input('value'); 


       if(SocialSettings::where_not_null($id)){

           $socialInsert->$id = $value;
           $socialInsert->save();
   }else{
        $socialCheck->update($value);

   }

    }

Answer 1

以下是类似问题的例子

SELECT ShopOrder.OrderDate
, Book.BookID
, Book.title
, COUNT(ShopOrder.ShopOrderID) AS "Total number of order"
, SUM (Orderline.Quantity) AS "Total quantity"
, Orderline.UnitSellingPrice * Orderline.Quantity  AS "Total order value"
, book.Price * Orderline.Quantity AS "Total retail value" 
      FROM ShopOrder
      JOIN Orderline
        ON Orderline.ShopOrderID = ShopOrder.ShopOrderID
      JOIN Book
        ON Book.BookID = Orderline.BookID 
      JOIN Publisher
        ON Publisher.PublisherID = Book.PublisherID
     WHERE Publisher.name = 'Addison Wesley'
     GROUP 
        BY ShopOrder.OrderDate
, Book.BookID
, Book.title
, Orderline.UnitSellingPrice
, Orderline.Quantity, book.Price 
, Orderline.Quantity, ShopOrder.ShopOrderID
     ORDER
        BY ShopOrder.OrderDate

喜欢这个

$user = User::firstOrNew(array('name' => Input::get('name')));
$user->foo = Input::get('foo');
$user->save();

firstOrNew Laravel API DOCS

如果存在数据，则更新db字段

1 个答案: